Darren drives to school in rush hour traffic and averages 32 mph. he returns home in mid-afternoon when there is less traffic and averages 48 mph. what is the distance between his home and school if the total traveling time is 1 hr 15 min?

d = distance between home and school

v₁ = average speed while going from home to school = 32 mph

t₁ = time taken to travel distance going from home to school

time taken to travel distance going from home to school is given as

t₁ = d/v₁

t₁ = d/32                                  eq-1

v₂ = average speed while going from school to home = 48 mph

t₂ = time taken to travel distance going from school to home

time taken to travel distance going from school to home is given as

t₂ = d/v₂

t₂ = d/48                                                     eq-2

t = total time of travel = 1 hour 15 min = 1.25 hour

total time of travel is given as

t = t₁ + t₂

using eq-1  and eq-2

1.25 = (d/32) + (d/48)

d = 24 miles

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap